Lm317 battery charger

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Hi Spec,
Sad to say, I can't get your design to work properly. I'm sure I have it all put together correctly and I've rechecked it several times. The red light never comes on. The 500R pot doesn't change anything. I tried on a 12v battery as it was designed for but I just can't get it to go.
 
ronsimpson said:
If the battery is already charged the red light will not come on.
Click to expand...
It's not! I even took it apart off the breadboard and put it back together from a different direction, just in case I screwed something up and missed it. The components the second time around were new again. It functioned the same way. I'll check it again if everyone is sure it should work.
 
Are you getting "14.7V"? What is there? I am trying to find out if the LM317 is working?
Is there 1.25 volts across "120R"?
 
ronsimpson said:
Are you getting "14.7V"? What is there? I am trying to find out if the LM317 is working?
Is there 1.25 volts across "120R"?
Click to expand...
Hi,
The lm317 is working. I can adjust the voltage over a wide range. Also there is about 10v at the collector of the npn transistor .18v on the base, 9.9v between the collector and base and .18v between the base and emitter. I don't think it's biased but I can't quite make out why. I'm a bit weak with transistors, sorry.
 
When current flows in the battery there is the same current in the "33R" resistor.
It takes (0.03 amps approx) to make one volt.
Please adjust the "500R" to zero ohms.
A transistor needs about 0.6 volts B to E to turn on.
"500R" and "330R" makes a voltage divider.
>>If "500R" is set to zero then there is 0.6 volts across "33R" the transistor turns on and tells the LM317 to reduce the voltage.
>>If "500R" is set to 330 ohms or about half way it makes a 2:1 divider. It will take 1.2 volts on "33R" to turn on the transistor.
>>If "500R"=500 it takes even more voltage (current) to turn on the transistor.
What is the voltage across "33R"?
 
Hi BM,

Oh dear, those are nightmare words for a paper designer.

I will have a look at the schematic and see if there is a problem.

With any feedback system there is always the danger of oscillation. Can you post an image of the circuit physical layout?

spec
 
Hi Ron,

Thanks for your help in fault-finding the circuit of post #8 - much appreciated.

spec
 
Hi BM,

Can you put a wire link across the 33 Ohm resistor and make sure that the battery charging voltage can be made to increase by adjusting the 5K Ohm potentiometer? The red charging LED will not illuminate.

Thanks

spec
 
Hi BM,

You are in Vancouver BC, so your mains supply is 120V RMS 60Hz. Is that correct?

With a 15V 0V 15V transformer, and two diode rectifier, the off-load voltage across the 470uF reservoir capacitor will be 20V with zero volts ripple. But, as the current drain increases, not only will the ripple voltage increase, but also the transformer secondary voltage will drop.

The peak-to-peak ripple voltage can be calculated from Vripple (p/p) = ( I * 8.3) / C
Where:
I = current drain in Amps
C= value of reservoir capacitor in uF

The other point is that the reservoir capacitor needs at be a low ESR high ripple current type or the rectifier performance will be degraded.

A further point is to take into account is the LM317 drop-out voltage, so the voltage between the input and output pin of the LM317 should never drop to less than 3V, even momentarily.

In view of the above, it would be worthwhile checking how your rectifier circuit is performing.

spec
 
Last edited:
spec said:
Hi BM,

Can you put a wire link across the 33 Ohm resistor and make sure that the battery charging voltage can be made to increase by adjusting the 5K Ohm potentiometer? The red charging LED will not illuminate.

Thanks

spec
Click to expand...
Hi,
So the voltage is adjustable as it is now. I don't need to short the 33R to get it to adjust. The voltage across the 33R is just .2v at 20VDC in. The circuit works the same with or without the 500r pot, 47r,330r,33R, red led, and the transistor. I am using a power supply for the test at 20VDC. So I've not included the 2 1n4007s of the mains. However if I remove only the 33r instead of shorting it, the 500r affects the current/voltage and will make the red led come on with low/no current and off with high current.
I'm not sure how it will react with an uncharged battery now because all this testing has charged the batteries I have so I need to discharge to carry on!
I'm out now so I'll try and get a pic of the breadboard layout for you later.
 
Hi BM,

The circuit sounds like it is working as it should.

After what you said, I now realise that the red charge LED will only illuminate when the circuit is in the constant current mode. With a light charge current, when the transistor will not be conducting, the LED will not illuminate. That is why removing the 33 Ohm resistor turns the red LED on when only a light current is flowing into the battery.

All the same, the battery charging function will be correct.

Can you set the current potentiometer to zero Ohms and put a discharged battery on the charger. And also connect a 470 Ohm resistor across the output. In both cases the red LED should illuminate.

spec
 
ronsimpson said:
This makes me think the battery current is very small. (Why? that is the question) You are probably right in that the battery is charged.
Click to expand...

Yes, that is correct and means a current of only 0.2V/33 Ohms = 6mA. But the absolute minimum current through the 33 Ohm resistor is 1.25V/120 Ohms =10.4 mili Amps.

spec
 
Hi again BM,

The way to set the circuit up for charging a 12v lead acid battery is to turn the current resistor (500 Ohms) to maximum resistance. Then connect a 470 Ohm resistor across the output of the charger (with no battery connected) and adjust the voltage resistor (5K) for 14.1V (2.35V per cell) at the charger output. The voltage resistor should not be adjusted again after that.

You can then set the current resistor to the maximum constant current that you want to charge the battery with, normally capacity/10. So, if for example, you had a 500mA Hr battery you would normally charge at 50mA for a good charge and a good battery life.

spec
 
Last edited:
Ok, I'll try it when I get home. A least one of those batteries should be dead again. Just to be clear though, when I first said it wasn't working the battery was not charged so maybe I had done something wrong at that time. Thanks for all the help so far, I learn a lot doing this kind of project. Eventhough I know quite a few of the principles involved, they don't always click together until something goes wrong in a project and it has to be figured out. Just the way I learn I guess.
 
Hi,

Here are two charger circuits for 4.2v that have worked well in the past for various people.
Note you can change the output voltage with one resistor and supply a higher voltage input.
The transistor Q1 circuit can be used in either circuit to limit current which eliminates the need for R1 the 30 ohm resistor in the LM317 circuit.
Cant get any simpler than that.
Also, the LM317 can be used in either circuit.
You can see from the dates these circuits have been around for a long time.
 

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Spec, Look at these two circuits. (Thanks MrAl for the second one) It will take 2 seconds to see the error in the first one.

 
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